84. Largest Rectangle in Histogram

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

Example:

Input: [2,1,5,6,2,3]
Output: 10

my code:(very inefficient)

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int len = heights.size();
        int res = 0;
        for (int i = 0; i < len; ++i) {
            int nums = 1;
            for (int j = i+1; j < len; ++j) {
                if (heights[j] >= heights[i])
                    nums++;
                else 
                    break;
            }
            for (int k = i-1; k >= 0; --k) {
                if (heights[k] >= heights[i])
                    nums++;
                else
                    break;
            }
            res = max(res, heights[i]*nums);
        }
        return res;
    }
};

Runtime: 1664 ms, faster than 1.28% of C++ online submissions for Largest Rectangle in Histogram.

height efficient code:

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int len = heights.size();
        int res = 0;
        stack<int> s;
        heights.push_back(0);
        for (int i = 0; i <= len; ++i) {
            int h = i==len ? 0 : heights[i];
            while (!s.empty() && h < heights[s.top()]) {
                int height = heights[s.top()];
                s.pop();
                int start = s.empty() ? -1 : s.top();
                int nums = i - start - 1;
                res = max(res, height*nums);
            }
            s.push(i);
        }
        
        return res;
    }
};

Runtime: 12 ms, faster than 45.96% of C++ online submissions for Largest Rectangle in Histogram.

 

stack中的数据（索引）升序进行排列，遇到小于前面的数开始计算，并将stack中比当前元素大的数pop。